Leftover Hash Lemma and Its Applications

Leftover Hash Lemma and Its Applications

Leftover Hash Lemma and its Applications rkm0959 (Gyumin Roh) October 18th Outline Introduction and Motivation Bunch of Terminology The Leftover Hash Lemma Application 1: Random Number Generation Application 2: Quantum Key Distribution Application 3: Learning With Errors Further Works References Table of Contents Introduction and Motivation Bunch of Terminology The Leftover Hash Lemma Application 1: Random Number Generation Application 2: Quantum Key Distribution Application 3: Learning With Errors Further Works References I Suppose I have a secret bit string of length 8, b1b2b3 ··· b8. I It doesn’t serve real meaning, just for password purposes. I You do somehow know that b1 6= b2, b2 = b3, b4 = 1. I You want to find my bit string by brute-force. I How many possibilities do you need to check? Introduction I Let’s begin with a simple puzzle :) I How many possibilities do you need to check? Introduction I Let’s begin with a simple puzzle :) I Suppose I have a secret bit string of length 8, b1b2b3 ··· b8. I It doesn’t serve real meaning, just for password purposes. I You do somehow know that b1 6= b2, b2 = b3, b4 = 1. I You want to find my bit string by brute-force. Introduction I Let’s begin with a simple puzzle :) I Suppose I have a secret bit string of length 8, b1b2b3 ··· b8. I It doesn’t serve real meaning, just for password purposes. I You do somehow know that b1 6= b2, b2 = b3, b4 = 1. I You want to find my bit string by brute-force. I How many possibilities do you need to check? I Here’s an intuitive explanation. I There are 8 unknown bits. I There are 3 bits of information. I Therefore, the answer is 28−3 = 32. Introduction I It’s not hard to see the answer is 32. Introduction I It’s not hard to see the answer is 32. I Here’s an intuitive explanation. I There are 8 unknown bits. I There are 3 bits of information. I Therefore, the answer is 28−3 = 32. I Case 1: I know what information you have. I Just change my secret to b1b5b6b7b8. I Now you know nothing, and there are 32 possibilities. I Case 2: I do not know what information you have. I Now what? I This is where Leftover Hash Lemma comes in! Introduction I I feel kinda bad that you have some information on my secret. I I think I should be able to "compress" my secret? I Goal: You have no information about my secret. I Case 2: I do not know what information you have. I Now what? I This is where Leftover Hash Lemma comes in! Introduction I I feel kinda bad that you have some information on my secret. I I think I should be able to "compress" my secret? I Goal: You have no information about my secret. I Case 1: I know what information you have. I Just change my secret to b1b5b6b7b8. I Now you know nothing, and there are 32 possibilities. I This is where Leftover Hash Lemma comes in! Introduction I I feel kinda bad that you have some information on my secret. I I think I should be able to "compress" my secret? I Goal: You have no information about my secret. I Case 1: I know what information you have. I Just change my secret to b1b5b6b7b8. I Now you know nothing, and there are 32 possibilities. I Case 2: I do not know what information you have. I Now what? Introduction I I feel kinda bad that you have some information on my secret. I I think I should be able to "compress" my secret? I Goal: You have no information about my secret. I Case 1: I know what information you have. I Just change my secret to b1b5b6b7b8. I Now you know nothing, and there are 32 possibilities. I Case 2: I do not know what information you have. I Now what? I This is where Leftover Hash Lemma comes in! I A: partial leakage of secret information can be critical! I A: brute-force is not the only possible attack in cryptography. I A: here’s a very good example in RSA. Textbook RSA I Public Key: N and e with gcd(e; φ(N)) = 1. I p; q are large primes, so factorization of N = pq is hard. I Private Key: d such that ed ≡ 1 (mod φ(N)). I Encryption of m: c = me (mod N). I Decryption of c: m = cd (mod N). I Factorization of N is equivalent to deriving d. I Details: rkm0959.tistory.com/131 Motivation I Q: Why do you feel the need to compress anyway? I Q: Amount of brute-force remains the same. I Q: Is rkm0959 delusional? Textbook RSA I Public Key: N and e with gcd(e; φ(N)) = 1. I p; q are large primes, so factorization of N = pq is hard. I Private Key: d such that ed ≡ 1 (mod φ(N)). I Encryption of m: c = me (mod N). I Decryption of c: m = cd (mod N). I Factorization of N is equivalent to deriving d. I Details: rkm0959.tistory.com/131 Motivation I Q: Why do you feel the need to compress anyway? I Q: Amount of brute-force remains the same. I Q: Is rkm0959 delusional? I A: partial leakage of secret information can be critical! I A: brute-force is not the only possible attack in cryptography. I A: here’s a very good example in RSA. Motivation I Q: Why do you feel the need to compress anyway? I Q: Amount of brute-force remains the same. I Q: Is rkm0959 delusional? I A: partial leakage of secret information can be critical! I A: brute-force is not the only possible attack in cryptography. I A: here’s a very good example in RSA. Textbook RSA I Public Key: N and e with gcd(e; φ(N)) = 1. I p; q are large primes, so factorization of N = pq is hard. I Private Key: d such that ed ≡ 1 (mod φ(N)). I Encryption of m: c = me (mod N). I Decryption of c: m = cd (mod N). I Factorization of N is equivalent to deriving d. I Details: rkm0959.tistory.com/131 Motivation Theorem (Partial Key Exposure : Coppersmith) Let N = pq be an n-bit RSA modulus. Then given the n=4 least significant bits of p or the n=4 most significant bits of p, one can efficiently factor N. Same applies for q. Theorem (Partial Key Exposure : Boneh, Durfee, Frank) Let hN; di be a private RSA key in which N is n bits long. Given the n=4 least significant bits of d, one can reconstruct all of d in time linear in e log e. Note that e = 216 + 1 is usually used. I A: First, generation of "truly random" bits is a hard task. I A: Also, plz just wait until applications section :P I Also, it’s not bad to be extra cautious in cryptography :) I Cryptographers have very high standards in their schemes. I Now that we have all the motivation, we begin. Motivation I Q: Wait, why not just generate a new random secret? I Also, it’s not bad to be extra cautious in cryptography :) I Cryptographers have very high standards in their schemes. I Now that we have all the motivation, we begin. Motivation I Q: Wait, why not just generate a new random secret? I A: First, generation of "truly random" bits is a hard task. I A: Also, plz just wait until applications section :P Motivation I Q: Wait, why not just generate a new random secret? I A: First, generation of "truly random" bits is a hard task. I A: Also, plz just wait until applications section :P I Also, it’s not bad to be extra cautious in cryptography :) I Cryptographers have very high standards in their schemes. I Now that we have all the motivation, we begin. Table of Contents Introduction and Motivation Bunch of Terminology The Leftover Hash Lemma Application 1: Random Number Generation Application 2: Quantum Key Distribution Application 3: Learning With Errors Further Works References I Then we will transform them into mathematical ones! I You have some information on my secret. I You have some probability distribution over my secret. I I have no idea what information you have. I I have no idea what your distribution is. I At least I hope/know that you have no good guesses. I max Pyou(secret = X ) should be small. X I I want to change my secret so that you have no info on it. I I want your distribution to be close to uniform. What Do We Want? I Let’s briefly summarize my goal, in simpler terms. I You have some information on my secret. I You have some probability distribution over my secret. I I have no idea what information you have. I I have no idea what your distribution is. I At least I hope/know that you have no good guesses. I max Pyou(secret = X ) should be small. X I I want to change my secret so that you have no info on it. I I want your distribution to be close to uniform. What Do We Want? I Let’s briefly summarize my goal, in simpler terms. I Then we will transform them into mathematical ones! I You have some probability distribution over my secret. I I have no idea what your distribution is.

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